Find the critical points and inflection points of the function f (x) = x4 -2x2 (with domain
the set of all real numbers). Which of the critical points are local minima? local
maxima? Is there a global minimum or maximum?
We first calculate the derivatives of the function:
4x3 - 4x
4(x + 1)x(x - 1)
12x2 - 4
4(3x2 - 1)
We see that f'(x) = 0 when x = - 1, 0, or 1, so these are the three critical points of
f. We calculate the second derivatives at these points:
so by the second derivative test, f has local minima at -1 and 1 and a local maximum
at 0. Substituting back into the original function yields
f (- 1)
so f attains its global minimum of -1 at x = ±1. It is clear from the graph of f
that it has no global maximum.
To find the points of inflection, we solve f''(x) = 0, or 12x2 - 4 = 0, which has
solutions x = ±1/3) ±0.58. Referring once again to the graph of f,
we can check that the concavity does indeed change at these x-values.
Find the inflection points of f (x) = e-x2. (This famous function
is called the gaussian.)
We compute the derivatives:
(- 2x)(- 2xe-x2) + (- 2)(e-x2)
Solving f''(x) = 0 for x, we get the inflection points x = ±1/ 0.71.
This is reasonable when one considers the graph of f.
Find a function f (x) with inflection points at x = 1 and x = 2.
We need f''(1) = f''(2) = 0, so we might as well let
f''(x) = (x - 1)(x - 2) = x2 - 3x + 2
We search for a function with this derivative, ultimately obtaining
f'(x) = - + 2x
Similarly, a function with this derivative is given by